Understanding Double Displacement Reactions
This reaction is a classic example of a double displacement reaction. In this type of reaction, the positive and negative ions of two reactants switch partners to form two new products.
The Reaction
The general form of the reaction is:
PbSO₄ (aq) + Sn(ClO₃)₂ (aq) → ? + ?
Predicting the Products
1. Identify the ions:
- Lead sulfate (PbSO₄) dissociates into Pb²⁺ (lead(II) ions) and SO₄²⁻ (sulfate ions).
- Tin chlorate (Sn(ClO₃)₂) dissociates into Sn²⁺ (tin(II) ions) and ClO₃⁻ (chlorate ions).
2. Switch partners:
- The lead(II) ions (Pb²⁺) will combine with the chlorate ions (ClO₃⁻) to form lead(II) chlorate (Pb(ClO₃)₂).
- The tin(II) ions (Sn²⁺) will combine with the sulfate ions (SO₄²⁻) to form tin(II) sulfate (SnSO₄).
The Balanced Chemical Equation
The balanced chemical equation for the reaction is:
PbSO₄ (aq) + Sn(ClO₃)₂ (aq) → Pb(ClO₃)₂ (aq) + SnSO₄ (s)
Important Note: Tin(II) sulfate (SnSO₄) is insoluble in water, so it will precipitate out of the solution as a solid.
Therefore, the newly formed compounds are lead(II) chlorate (Pb(ClO₃)₂), which remains dissolved in the solution, and tin(II) sulfate (SnSO₄), which precipitates out as a solid.
